Student Seminar: Classical and Quantum Integrable Systems

provided W A n + W D n = 0 for all n, which means that
(
λ n + i ) L ∏ M
λ n − λ j − i
(
= λ n − i 2 λ
j=1 n − λ j 2
j≠n
We write the last equations in the form
( ) L λn + i/2
=
λ n − i/2
M∏
j=1
j≠n
) L M
∏
j=1
j≠n
λ n − λ j + i
λ n − λ j − i .
λ n − λ j + i
λ n − λ j
.
These are the Bethe equations. Introducing λ j = cot p j the Bethe equations take
precisely the same form as derived in the coordinate Bethe ansatz approach:
e ip iL =
M∏
S(p i , p j ) .
j=1
j≠i
Note that the parametrization λ j = cot p j has a singularity at k j = 0. From the
experience with the coordinate Bethe ansatz we know that all the eigenvectors for
which k j ≠ 0 are the highest weight states of the global spin algebra su(2). Thus, we
expect that the eigenvectors obtained in the algebraic Bethe ansatz approach have
the same property. Now we are going to investigate this issue in mode detail.
Realization of the symmetry algebra. Let us consider the fundamental commutation
relations in the limitimg case µ → ∞. We get
(
(λ − µ) + i 2 (I a ⊗ I b + ∑ α
) (
σa α ⊗ σb α ) T a (λ) µ L + iµ ∑ )
L−1 Sn α ⊗ σb α + · · · =
n,α
=
(
µ L + iµ ∑ ) (
L−1 Sn α ⊗ σb α + · · · T a (λ) (λ − µ) + i 2 (I a ⊗ I b + ∑
n,α
α
σ α a ⊗ σ α b ) .
The leading term of the order µ L+1 cancel out. The subleading term of order µ L
gives
−iT a (λ) ∑ Sn α ⊗ σb α + i 2 T a(λ) + i ( ∑ )
σa α ⊗ σb
α T a (λ) =
2
n,α
α
= i 2 T a(λ) + i (
2 T ∑ )
a(λ) σa α ⊗ σb
α − i ∑ Sn α ⊗ σb α T a (λ) .
α
n,α
Simplifying we get
∑
[T a (λ), S α + 1 2 σα a ] ⊗ σb α = 0 .
α
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